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Q. The work done by an uniform magnetic field, on a moving charge is

VITEEEVITEEE 2019

Solution:

Force on moving charge while moving in magnetic field is; $\vec{F}=q\left(\vec{v}\times\vec{B} \right)$ where $\vec{F}$ is perpendicular to $\vec{v}$
Work done/sec $=\vec{F},\vec{v}=Fv\,cos\,90ยบ=0$